A082732 -id:A082732 - cp's OEIS frontend

A144802 Decimal expansion of the constant 1.880678543683078094492191765... arising in A082732.

1, 8, 8, 0, 6, 7, 8, 5, 4, 3, 6, 8, 3, 0, 7, 8, 0, 9, 4, 4, 9, 2, 1, 9, 1, 7, 6, 5, 0, 1, 2, 7, 5, 0, 3, 5, 6, 2, 6, 3, 0, 6, 1, 7, 5, 6, 3, 2, 3, 6, 3, 0, 1, 9, 6, 9, 0, 4, 7, 9, 9, 5, 9, 5, 3, 3, 9, 1, 4, 7, 9, 8, 7, 1, 7, 6, 9, 5, 3, 9, 5, 2, 0, 4, 0, 8, 7, 3, 5, 8, 0, 9, 0, 8, 7, 4, 1, 9, 4, 1, 2, 4, 5, 0, 3

Offset: 1

Artur Jasinski, Sep 22 2008

Cf. A082732, A076393, A144803, A144804, A144805, A144806, A144807, A144808, A144809, A144810, A144811, A144812.

c=1.880678543683078094492191765...; A082732(n+2) = c^(2^n).

A000058 Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(0) = 2.

Original entry on oeis.org

2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, 12864938683278671740537145998360961546653259485195807

Offset: 0

N. J. A. Sloane

Also called Euclid numbers, because a(n) = a(0)*a(1)*...*a(n-1) + 1 for n>0, with a(0)=2. - Jonathan Sondow, Jan 26 2014
Another version of this sequence is given by A129871, which starts with 1, 2, 3, 7, 43, 1807, ... .
The greedy Egyptian representation of 1 is 1 = 1/2 + 1/3 + 1/7 + 1/43 + 1/1807 + ... .
Take a square. Divide it into 2 equal rectangles by drawing a horizontal line. Divide the upper rectangle into 2 squares. Now you can divide the lower one into another 2 squares, but instead of doing so draw a horizontal line below the first one so you obtain a (2+1 = 3) X 1 rectangle which can be divided in 3 squares. Now you have a 6 X 1 rectangle at the bottom. Instead of dividing it into 6 squares, draw another horizontal line so you obtain a (6+1 = 7) X 1 rectangle and a 42 X 1 rectangle left, etc. - Néstor Romeral Andrés, Oct 29 2001
More generally one may define f(1) = x_1, f(2) = x_2, ..., f(k) = x_k, f(n) = f(1)*...*f(n-1)+1 for n > k and natural numbers x_i (i = 1, ..., k) which satisfy gcd(x_i, x_j) = 1 for i <> j. By definition of the sequence we have that for each pair of numbers x, y from the sequence gcd(x, y) = 1. An interesting property of a(n) is that for n >= 2, 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) = (a(n)-2)/(a(n)-1). Thus we can also write a(n) = (1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) - 2 )/( 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n-1) - 1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), May 10 2001; [corrected by Michel Marcus, Mar 27 2019]
A greedy sequence: a(n+1) is the smallest integer > a(n) such that 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(n+1) doesn't exceed 1. The sequence gives infinitely many ways of writing 1 as the sum of Egyptian fractions: Cut the sequence anywhere and decrement the last element. 1 = 1/2 + 1/3 + 1/6 = 1/2 + 1/3 + 1/7 + 1/42 = 1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = ... . - Ulrich Schimke, Nov 17 2002; [corrected by Michel Marcus, Mar 27 2019]
Consider the mapping f(a/b) = (a^3 + b)/(a + b^3). Starting with a = 1, b = 2 and carrying out this mapping repeatedly on each new (reduced) rational number gives 1/2, 1/3, 4/28 = 1/7, 8/344 = 1/43, ..., i.e., 1/2, 1/3, 1/7, 1/43, 1/1807, ... . Sequence contains the denominators. Also the sum of the series converges to 1. - Amarnath Murthy, Mar 22 2003
a(1) = 2, then the smallest number == 1 (mod all previous terms). a(2n+6) == 443 (mod 1000) and a(2n+7) == 807 (mod 1000). - Amarnath Murthy, Sep 24 2003
An infinite coprime sequence defined by recursion.
Apart from the initial 2, a subsequence of A002061. It follows that no term is a square.
It appears that a(k)^2 + 1 divides a(k+1)^2 + 1. - David W. Wilson, May 30 2004. This is true since a(k+1)^2 + 1 = (a(k)^2 - a(k) + 1)^2 +1 = (a(k)^2-2*a(k)+2)*(a(k)^2 + 1) (a(k+1)=a(k)^2-a(k)+1 by definition). - Pab Ter (pabrlos(AT)yahoo.com), May 31 2004
In general, for any m > 0 coprime to a(0), the sequence a(n+1) = a(n)^2 - m*a(n) + m is infinite coprime (Mohanty). This sequence has (m,a(0))=(1,2); (2,3) is A000215; (1,4) is A082732; (3,4) is A000289; (4,5) is A000324.
Any prime factor of a(n) has -3 as its quadratic residue (Granville, exercise 1.2.3c in Pollack).
Note that values need not be prime, the first composites being 1807 = 13 * 139 and 10650056950807 = 547 * 19569939581. - Jonathan Vos Post, Aug 03 2008
If one takes any subset of the sequence comprising the reciprocals of the first n terms, with the condition that the first term is negated, then this subset has the property that the sum of its elements equals the product of its elements. Thus -1/2 = -1/2, -1/2 + 1/3 = -1/2 * 1/3, -1/2 + 1/3 + 1/7 = -1/2 * 1/3 * 1/7, -1/2 + 1/3 + 1/7 + 1/43 = -1/2 * 1/3 * 1/7 * 1/43, and so on. - Nick McClendon, May 14 2009
(a(n) + a(n+1)) divides a(n)*a(n+1)-1 because a(n)*a(n+1) - 1 = a(n)*(a(n)^2 - a(n) + 1) - 1 = a(n)^3 - a(n)^2 + a(n) - 1 = (a(n)^2 + 1)*(a(n) - 1) = (a(n) + a(n)^2 - a(n) + 1)*(a(n) - 1) = (a(n) + a(n+1))*(a(n) - 1). - Mohamed Bouhamida, Aug 29 2009
This sequence is also related to the short side (or hypotenuse) of adjacent right triangles, (3, 4, 5), (5, 12, 13), (13, 84, 85), ... by A053630(n) = 2*a(n) - 1. - Yuksel Yildirim, Jan 01 2013, edited by M. F. Hasler, May 19 2017
For n >= 4, a(n) mod 3000 alternates between 1807 and 2443. - Robert Israel, Jan 18 2015
The set of prime factors of a(n)'s is thin in the set of primes. Indeed, Odoni showed that the number of primes below x dividing some a(n) is O(x/(log x log log log x)). - Tomohiro Yamada, Jun 25 2018
Sylvester numbers when reduced modulo 864 form the 24-term arithmetic progression 7, 43, 79, 115, 151, 187, 223, 259, 295, 331, ..., 763, 799, 835 which repeats itself until infinity. This was first noticed in March 2018 and follows from the work of Sondow and MacMillan (2017) regarding primary pseudoperfect numbers which similarly form an arithmetic progression when reduced modulo 288. Giuga numbers also form a sequence resembling an arithmetic progression when reduced modulo 288. - Mehran Derakhshandeh, Apr 26 2019
Named after the English mathematician James Joseph Sylvester (1814-1897). - Amiram Eldar, Mar 09 2024
Guy askes if it is an irrationality sequence (see Guy, 1981). - Stefano Spezia, Oct 13 2024

			a(0)=2, a(1) = 2+1 = 3, a(2) = 2*3 + 1 = 7, a(3) = 2*3*7 + 1 = 43.

Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 6.7.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994.
Richard K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E24.
Richard K. Guy and Richard Nowakowski, Discovering primes with Euclid. Delta, Vol. 5 (1975), pp. 49-63.
Amarnath Murthy, Smarandache Reciprocal partition of unity sets and sequences, Smarandache Notions Journal, Vol. 11, 1-2-3, Spring 2000.
Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.1.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..12
David Adjiashvili, Sandro Bosio and Robert Weismantel, Dynamic Combinatorial Optimization: a complexity and approximability study, 2012.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.
Gennady Bachman and Troy Kessler, On divisibility properties of certain multinomial coefficients—II, Journal of Number Theory, Volume 106, Issue 1, May 2004, Pages 1-12.
Andreas Bäuerle, Sharp volume and multiplicity bounds for Fano simplices, arXiv:2308.12719 [math.CO], 2023.
Kevin S. Brown, Odd, Greedy and Stubborn (Unit Fractions).
Eunice Y. S. Chan and Robert M. Corless, Minimal Height Companion Matrices for Euclid Polynomials, Mathematics in Computer Science, Vol. 13, No. 1-2 (2019), pp. 41-56, arXiv preprint, arXiv:1712.04405 [math.NA], 2017.
Hung Viet Chu, A Threshold for the Best Two-term Underapproximation by the Greedy Algorithm, arXiv:2306.12564 [math.NT], 2023.
Matthew Brendan Crawford, On the Number of Representations of One as the Sum of Unit Fractions, Master's Thesis, Virginia Polytechnic Institute and State University (2019).
D. R. Curtiss, On Kellogg's Diophantine problem, Amer. Math. Monthly, Vol. 29, No. 10 (1922), pp. 380-387.
Mehran Derakhshandeh, Why do Sylvester numbers, when reduced modulo 864, form an arithmetic progression 7,43,79,115,151,187,223,...?
Paul Erdős and E. G. Straus, On the Irrationality of Certain Ahmes Series, J. Indian Math. Soc. (N.S.), 27(1964), pp. 129-133.
Steven Finch, Exercises in Iterational Asymptotics, arXiv:2411.16062 [math.NT], 2024. See p. 10.
Solomon W. Golomb, On the sum of the reciprocals of the Fermat numbers and related irrationalities, Canad. J. Math., 15 (1963), 475-478.
Solomon W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly 70 (1963), 403-405.
Richard K. Guy and Richard Nowakowski, Discovering primes with Euclid, Research Paper No. 260 (Nov 1974), The University of Calgary Department of Mathematics, Statistics and Computing Science.
János Kollár, Which powers of holomorphic functions are integrable?, arXiv:0805.0756 [math.AG], 2008.
E. Lemoine, Sur la décomposition d'un nombre en ses carrés maxima, Assoc. Française pour L'Avancement des Sciences (1896), 73-77.
Zheng Li and Quanyu Tang, On a conjecture of Erdős and Graham about the Sylvester's sequence, arXiv:2503.12277 [math.NT], 2025. See p. 2.
Nick Lord, A uniform construction of some infinite coprime sequences, The Mathematical Gazette, vol. 92, no. 523, March 2008, pp.66-70.
Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012. - _N. J. A. Sloane_, Jun 13 2012
Melvyn B. Nathanson, Underapproximation by Egyptian fractions, arXiv:2202.00191 [math.NT], 2022.
Benjamin Nill, Volume and lattice points of reflexive simplices, Discrete & Computational Geometry, Vol. 37, No. 2 (2007), pp. 301-320, arXiv preprint, arXiv:math/0412480 [math.AG], 2004-2007.
R. W. K. Odoni, On the prime divisors of the sequence w_{n+1}=1+w_1 ... w_n, J. London Math. Soc. 32 (1985), 1-11.
Michael Penn, An intriguing integer sequence — Sylvester’s Sequence, YouTube video (2022).
Simon Plouffe, A set of formulas for primes, arXiv:1901.01849 [math.NT], 2019.
Paul Pollack, Analytic and Combinatorial Number Theory Course Notes, p. 5. [?Broken link]
Paul Pollack, Analytic and Combinatorial Number Theory Course Notes, p. 5.
Filip Saidak, A New Proof of Euclid's Theorem, Amer. Math. Monthly, Vol. 113, No. 10 (Dec., 2006), pp. 937-938.
N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]
Jonathan Sondow and Kieren MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdős-Moser equation, Amer. Math. Monthly, Vol. 124, No. 3 (2017), pp. 232-240, arXiv preprint, arXiv:math/1812.06566 [math.NT], 2018.
J. J. Sylvester, On a Point in the Theory of Vulgar Fractions, Amer. J. Math., Vol. 3, No. 4 (1880), pp. 332-335.
Burt Totaro, The ACC conjecture for log canonical thresholds, Séminaire Bourbaki no. 1025 (juin 2010).
Burt Totaro and Chengxi Wang, Varieties of general type with small volume, arXiv:2104.12200 [math.AG], 2021.
Akiyoshi Tsuchiya, The delta-vectors of reflexive polytopes and of the dual polytopes, Discrete Mathematics, Vol. 339, No. 10 (2016), pp. 2450-2456, arXiv preprint, arXiv:1411.2122 [math.CO], 2014-2016.
Stephan Wagner and Volker Ziegler, Irrationality of growth constants associated with polynomial recursions, arXiv:2004.09353 [math.NT], 2020.
Eric Weisstein's World of Mathematics, Quadratic Recurrence Equation.
Eric Weisstein's World of Mathematics, Sylvester's Sequence.
Wikipedia, Sylvester's sequence.
Bowen Yao, A note on fraction decompositions of integers, The American Mathematical Monthly, 127(10), 928-932, Dec 2020.
Index entries for sequences of form a(n+1)=a(n)^2 + ....
Index entries for "core" sequences.

Cf. A005267, A000945, A000946, A005265, A005266, A075442, A007018, A014117, A054377, A002061, A118227, A126263, A007996 (primes dividing some term), A323605 (smallest prime divisors), A091335 (number of prime divisors), A129871 (a variant starting with 1).

a000058 0 = 2
a000058 n = a000058 m ^ 2 - a000058 m + 1 where m = n - 1
-- James Spahlinger, Oct 09 2012

a000058_list = iterate a002061 2  -- Reinhard Zumkeller, Dec 18 2013

a(n) = n == 0 ? BigInt(2) : a(n - 1)*(a(n - 1) - 1) + 1
[a(n) for n in 0:8] |> println # Peter Luschny, Dec 15 2020

A[0]:= 2:
for n from 1 to 12 do
A[n]:= A[n-1]^2 - A[n-1]+1
od:
seq(A[i],i=0..12); # Robert Israel, Jan 18 2015

a[0] = 2; a[n_] := a[n - 1]^2 - a[n - 1] + 1; Table[ a[ n ], {n, 0, 9} ]
NestList[#^2-#+1&,2,10] (* Harvey P. Dale, May 05 2013 *)
RecurrenceTable[{a[n + 1] == a[n]^2 - a[n] + 1, a[0] == 2}, a, {n, 0, 10}] (* Emanuele Munarini, Mar 30 2017 *)

a(n) := if n = 0 then 2 else a(n-1)^2-a(n-1)+1 $
makelist(a(n),n,0,8); /* Emanuele Munarini, Mar 23 2017 */

a(n)=if(n<1,2*(n>=0),1+a(n-1)*(a(n-1)-1))

A000058(n,p=2)={for(k=1,n,p=(p-1)*p+1);p} \\ give Mod(2,m) as 2nd arg to calculate a(n) mod m. - M. F. Hasler, Apr 25 2014

a=vector(20); a[1]=3; for(n=2, #a, a[n]=a[n-1]^2-a[n-1]+1); concat(2, a) \\ Altug Alkan, Apr 04 2018

A000058 = [2]
for n in range(1, 10):
    A000058.append(A000058[n-1]*(A000058[n-1]-1)+1)
# Chai Wah Wu, Aug 20 2014

a(n) = 1 + a(0)*a(1)*...*a(n-1).
a(n) = a(n-1)*(a(n-1)-1) + 1; Sum_{i>=0} 1/a(i) = 1. - Néstor Romeral Andrés, Oct 29 2001
Vardi showed that a(n) = floor(c^(2^(n+1)) + 1/2) where c = A076393 = 1.2640847353053011130795995... - Benoit Cloitre, Nov 06 2002 (But see the Aho-Sloane paper!)
a(n) = A007018(n+1)+1 = A007018(n+1)/A007018(n) [A007018 is a(n) = a(n-1)^2 + a(n-1), a(0)=1]. - Gerald McGarvey, Oct 11 2004
a(n) = sqrt(A174864(n+1)/A174864(n)). - Giovanni Teofilatto, Apr 02 2010
a(n) = A014117(n+1)+1 for n = 0,1,2,3,4; a(n) = A054377(n)+1 for n = 1,2,3,4. - Jonathan Sondow, Dec 07 2013
a(n) = f(1/(1-(1/a(0) + 1/a(1) + ... + 1/a(n-1)))) where f(x) is the smallest integer > x (see greedy algorithm above). - Robert FERREOL, Feb 22 2019
From Amiram Eldar, Oct 29 2020: (Start)
Sum_{n>=0} (-1)^n/(a(n)-1) = A118227.
Sum_{n>=0} (-1)^n/a(n) = 2 * A118227 - 1. (End)

A144780 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 6.

Original entry on oeis.org

6, 31, 931, 865831, 749662454731, 561993796032558961827631, 315837026779085485103718410756049100028793244531

Offset: 1

Artur Jasinski, Sep 21 2008

Hugo Pfoertner, Table of n, a(n) for n = 1..11
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144779, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144804.

a = {}; k = 6; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,6,10] (* Harvey P. Dale, Dec 19 2024 *)

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 6.

a(n) ~ c^(2^n) where is c is 2.350117384... (A144804).

a(8) moved to b-file by Hugo Pfoertner, Aug 30 2020

A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

Original entry on oeis.org

11, 111, 12211, 149096311, 22229709804712411, 494159998001727075769152612720511, 244194103625066907517263589918036880566782292998362610615987380611

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

For the "exact" formula, compare the Aho-Sloane reference in A000058. - N. J. A. Sloane, Apr 07 2014

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144808.

See A239900 for another version.

a = {}; r = 11; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

a(n) ~ c^(2^n) where c = 3.242214... (see A144808).

A144779 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

Original entry on oeis.org

5, 21, 421, 176821, 31265489221, 977530816197201697621, 955566496615167328821993756200407115362021, 913107329453384594090655605142589591944556891901674138343716072975722193082773842421

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

			a(0) = 4, a(1) = 4+1 = 5, a(2) = 4*5+1 = 21, a(3) = 4*5*21+1 = 421, a(4) = 4*5*21*421+1 = 176821, ... - _Philippe Deléham_, Apr 19 2013

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution, College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; k = 5; Do[AppendTo[a, k]; k = k^2 - k + 1, {n,1,10}]; a (* Artur Jasinski, Sep 21 2008 *)
NestList[#^2-#+1&,5,8] (* Harvey P. Dale, Jan 17 2012 *)

a(n) = round(2.127995907464107054577351...)^(2^n) = round(A144803^(2^n)). [corrected by Joerg Arndt, Jan 15 2021]

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

A144781 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.

Original entry on oeis.org

8, 57, 3193, 10192057, 103878015699193, 10790642145601683494645152057, 116437957914435303575899742229333045108448631998006179193, 13557798043283806950297045269968250387897834581711367551819275131055206893868524458302302046950954641412419952057

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144805.

a = {}; k = 8; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,8,10] (* Harvey P. Dale, Jan 29 2017 *)

a(n) ~ c^(2^n) where is c is 2.74167747444233776776... (A144805).

A144782 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

Original entry on oeis.org

9, 73, 5257, 27630793, 763460694178057, 582872231554839914154126117193, 339740038317718918529575265905277902175236102890836244082057

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; r = 9; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,9,10] (* Harvey P. Dale, Aug 31 2014 *)

a(n) ~ c^(2^n) with c = 2.918012...

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

A144783 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

Original entry on oeis.org

10, 91, 8191, 67084291, 4500302031888391, 20252718378218776104731448680491, 410172601707440572557971589875869064610540321970215293555320591, 168241563191450680898537024308131628447885486994777537422995633998657738457104605412468520116391629012196009150161991233268691

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; r = 10; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a (* or *)
Table[Round[3.08435104906918990233569320020272148875011089837398848476442237096569188195734783139337492942278549518507672786196650938869338548385641623^(2^n)], {n, 1, 8}] (* Artur Jasinski *)
NestList[#^2-#+1&,10,8] (* Harvey P. Dale, May 07 2017 *)

a(n) = round((3.08435104906918990233569320020272148875011089837398848476442237096569...)^(2^n)) = round(A144807^(2^n)). [corrected by Joerg Arndt, Jan 15 2021]

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

A144785 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 12.

Original entry on oeis.org

12, 133, 17557, 308230693, 95006159799029557, 9026170399758739819525199160586693, 81471752085480849000657595909467634426991447160798281416700808089557

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788

a = {}; r = 12; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a or Table[Round[3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n)], {n, 1, 8}] (*Artur Jasinski*)
NestList[#^2-#+1&,12,6] (* Harvey P. Dale, Jan 01 2016 *)

a(n) =3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n) a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11

A144786 If n is an oblong number A002378, then a(n)=a(j) where j is the number of oblong numbers in (0,n], otherwise a(n)=n.

Original entry on oeis.org

1, 1, 3, 4, 5, 1, 7, 8, 9, 10, 11, 3, 13, 14, 15, 16, 17, 18, 19, 4, 21, 22, 23, 24, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 1, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 8, 73, 74, 75, 76

Offset: 1

Artur Jasinski, Sep 22 2008, Sep 26 2008

As a motivation, consider the greedy decomposition of fractions 1/n into Egyptian fractions,
n=1: 2,3,7,43,1807,3263443,.. A000058
n=2: 3,7,43,1807,3263443,10650056950807,.. A000058
n=3: 4,13,157,24493,599882557,359859081592975693,.. A082732
n=4: 5,21,421,176821,31265489221,977530816197201697621,.. A144779
n=5: 6,31,931,865831,749662454731,561993796032558961827631,.. A144780
n=6: 7,43,1807,3263443,10650056950807,.. A000058
n=7: 8,57,3193,10192057,103878015699193,.. A144781
n=8: 9,73,5257,27630793,763460694178057,.. A144782
n=9: 10,91,8191,67084291,4500302031888391,.. A144783
n=10: 11,111,12211,149096311,22229709804712411,.. A144784
n=11: 12,133,17557,308230693,95006159799029557,.. A144785
n=12: 13,157,24493,599882557,.. A082732
k=13: 14,183,33307,1109322943,..
where the first few denominators of 1/n = 1/b(1)+1/b(2)+... have been tabulated.
For some sets of n, the list b(i) of denominators is essentially the same: consider for example A000058, which represents primarily n=1, then in truncated form also n=2, and then n=6, n=42 etc. Or consider A082732 which represents n=3, then in truncated form n=12, n=156 etc.
The OEIS sequence assigns the primary n to a(n). The interpretation of a(n) with ascending n is: n=1 is primary, a(1)=1.
Decomposition of n=2 is equivalent to n=1, a(2)=1. Cases n=3 to 5 are primary ("original", "new"), and a(n)=n in these cases. n=6 is not new but essentially the same Egyptian series as seen for n=1, so a(6)=1. Cases n=7 to n=11 are "new" sequences, again a(n)=n in these cases, but then n=12 is represented by A082732 as already seen for n=3, so a(12)=3.
Because the first denominator for the decomposition of 1/n is 1/(n+1), n+1 belongs to the sequence of denominators of the expansion of 1/a(n).
The sequences b(.) have recurrences which are essentially 1+b(n-1)*(b(n-1)-1), looking up the oblong number at the position of the previous b(.). This is the reason why reverse look-up of the n via A000194 (number of oblong numbers up to n) as used in the definition is equivalent to the assignment described above.

			n=1 is not in A002378, so a(1)=1.
n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0.
n=3 is not in A002378, so a(3)=3.
n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0.

Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786.

a(n) = a(A000194(n+1)) if n in A002378. a(n) = n if n in A078358.

a(57)=57 inserted, a(61)=61 corrected and better definition provided by Omar E. Pol, Dec 29 2008
I did some further editing of this entry, but many of the lines are still obscure. - N. J. A. Sloane, Dec 29 2008
Comments that connect to Egyptian fractions rephrased by R. J. Mathar, Oct 01 2009

cp's OEIS Frontend

A144802 Decimal expansion of the constant 1.880678543683078094492191765... arising in A082732.

Views

Author

Keywords

Crossrefs

Formula

A000058 Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(0) = 2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Haskell

Julia

Maple

Mathematica

Maxima

PARI

PARI

PARI

Python

Formula

A144780 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 6.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A144779 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A144781 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A144782 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A144783 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

Views

Author

Keywords